20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? ($ pK_a $ value of acetic acid is 4.75).

Question

Consider the reaction: $$ X_2Y(g) \rightleftharpoons X_2(g) + \frac{1}{2} Y_2(g) $$ The equation representing the correct relationship between the degree of dissociation $ x $ of $ X_2Y(g) $ with its equilibrium constant $ K_p $ is:

Answer 1

Step 1: Understanding the Concept:
When a weak acid is partially neutralized by a strong base, a buffer solution containing the weak acid and its conjugate base (salt) is formed. The pH of this buffer is calculated using the Henderson-Hasselbalch equation.
Step 2: Key Formula or Approach:
Neutralization reaction: $\text{CH}_3\text{COOH} + \text{NaOH} \to \text{CH}_3\text{COONa} + \text{H}_2\text{O}$.
Henderson-Hasselbalch equation: $\text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$.
Step 3: Detailed Explanation:
From the full neutralization data:
$20 \text{ mL}$ of acetic acid requires $28.4 \text{ mL}$ of $0.1 \text{ M}$ NaOH.
Milli-moles (mmol) of NaOH used for full neutralization = $M \times V = 0.1 \text{ mol/L} \times 28.4 \text{ mL} = 2.84 \text{ mmol}$.
Because NaOH and acetic acid react in a 1:1 molar ratio, the $20 \text{ mL}$ of acetic acid contains exactly $2.84 \text{ mmol}$ of acid.
Preparation of Solution (X):
We mix $20 \text{ mL}$ of this same acetic acid (containing $2.84 \text{ mmol}$) with $14.2 \text{ mL}$ of $0.1 \text{ M}$ NaOH.
Milli-moles of NaOH added = $0.1 \times 14.2 = 1.42 \text{ mmol}$.
The NaOH will react with the acetic acid:
Initial: Acid = $2.84 \text{ mmol}$, NaOH = $1.42 \text{ mmol}$, Salt = $0$
Reacts: $-1.42 \text{ mmol}$ acid, $-1.42 \text{ mmol}$ NaOH, $+1.42 \text{ mmol}$ Salt
Final: Acid = $2.84 - 1.42 = 1.42 \text{ mmol}$. NaOH = $0$. Salt = $1.42 \text{ mmol}$.
The resulting mixture is an acidic buffer containing equal amounts of the weak acid and its salt.
Apply the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \] Since they are in the same total volume, the ratio of concentrations equals the ratio of milli-moles:
\[ \text{pH} = 4.75 + \log \left( \frac{1.42}{1.42} \right) \] \[ \text{pH} = 4.75 + \log(1) = 4.75 + 0 = 4.75 \] Step 4: Final Answer:
The pH of solution (X) is 4.75.

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (\( pK_a \) value of acetic acid is 4.75).

The Correct Option is B

Solution and Explanation

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